Problem: What is the value of the following logarithm? $\log_{4} \left(\dfrac{1}{16}\right)$
Solution: If $b^y = x$ , then $\log_{b} x = y$ Therefore, we want to find the value $y$ such that $4^{y} = \dfrac{1}{16}$ In this case, $4^{-2} = \dfrac{1}{16}$, so $\log_{4} \left(\dfrac{1}{16}\right) = -2$.